CBSE Class 12 Mathematics Application of Derivatives VBQs Set A

BASIC CONCEPTS

Rate of Change: If a quantity \( y \) varies with another quantity \( x \), satisfying some rule \( y = f(x) \), then \( \frac{dy}{dx}\big|_{x = x_0} \) (or \( f'(x_0) \)) represents the rate of change of \( y \) with respect to \( x \) at \( x = x_0 \).

Differentials: Let \( y = f(x) \) be any function of \( x \) which is differentiable in \( (a, b) \). The derivative of this function at some point \( x \) of \( (a, b) \) is given by the relation
\( \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = f'(x) \)
\( \Rightarrow \frac{dy}{dx} = f'(x) \)
\( \Rightarrow dy = f'(x) dx \), where \( dy \) is called the differential of the function.

Note: Formula \( dy = f'(x) dx \) or \( \Delta y = f'(x) \Delta x \) is very useful in measuring the errors in the dependent variable for given error in independent variable.

• Absolute Error: The error \( \Delta x \) in \( x \) is called the absolute error.
• Relative Error: If \( \Delta x \) is error in \( x \) then ratio \( \frac{\Delta x}{x} \) is called the relative error.
• Percentage Error: If \( \frac{\Delta x}{x} \) is relative error, then \( \frac{\Delta x}{x} \times 100 \) is called percentage error in \( x \).

Tangents and Normals:

Slope of chord \( AB = \frac{f(a + \Delta x) - f(a)}{a + \Delta x - a} = \frac{f(a + \Delta x) - f(a)}{\Delta x} \)
Obviously, if \( \Delta x \to 0 \) comes very close to \( A \) and then chord \( AB \) becomes tangent at \( A \) i.e., \( x = a \).
i.e., slope of tangent at \( (a, f(a)) = \lim_{\Delta x \to 0} \frac{f(a + \Delta x) - f(a)}{\Delta x} = [f'(x)]_{(a, f(a))} = \left(\frac{dy}{dx}\right)_{(a, f(a))} \)
Hence, equation of tangent to the curve \( y = f(x) \) at the point \( (x_1, y_1) \) is given by
\( (y - y_1) = \left(\frac{dy}{dx}\right)_{(x_1, y_1)} (x - x_1) \) [ \(\because\) Equation of line is \( y - y_1 = m(x - x_1) \), where \( m \) is slope]

If \( \frac{dy}{dx} = \infty \) at the point \( P(x_1, y_1) \), then the tangent at \( P \) is parallel to y-axis and its equation is given by
\( \infty = \frac{y - y_1}{x - x_1} \Rightarrow x - x_1 = 0 \Rightarrow x = x_1 \)
[Note : In this case, \( \frac{dx}{dy} \) at \( P(x_1, y_1) = 0 \)]

If \( \frac{dy}{dx} = 0 \), at the point \( P(x_1, y_1) \), then the tangent at \( P \) is parallel to x-axis and equation is given as
\( 0 = \frac{y - y_1}{x - x_1} \Rightarrow y - y_1 = 0 \Rightarrow y = y_1 \)

Obviously, normal to the curve \( y = f(x) \) at \( P(x_1, y_1) \) is perpendicular to the tangent at \( P(x_1, y_1) \).
\(\therefore\) Slope of normal \( = \frac{-1}{\text{slope of tangent}} = \frac{-1}{\left(\frac{dy}{dx}\right)_{(x_1, y_1)}} \)
Hence, equation of normal to the curve \( y = f(x) \) at \( P(x_1, y_1) \) is
\( (y - y_1) = \frac{-1}{\left[\frac{dy}{dx}\right]_{(x_1, y_1)}} (x - x_1) \)

If \( \frac{dy}{dx} \) at the point \( P(x_1, y_1) \) is zero, then the equation of normal is \( x = x_1 \).

If \( \left(\frac{dy}{dx}\right) \) at the point \( (x_1, y_1) \) does not exist, then the equation of normal is \( y = y_1 \).

The angle \( \theta \) between two given curves \( y_1 = f_1(x) \) and \( y_2 = f_2(x) \) at a point \( (x_1, y_1) \) is given by
\( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \) where \( (x_1, y_1) \) is the point of intersection and \( m_1, m_2 \) are slopes of their tangents at point \( (x_1, y_1) \).

Note: The gradient of a curve at a point is defined as the slope of tangent to the curve at that point.

Nature of Function: To know the behaviour of a function in an interval, we study the properties of increasing or decreasing functions, maximum and minimum of the functions.

Monotonic Function: A function is said to be monotonic in an interval, if it is either increasing or decreasing in the given interval.

Increasing Function: A function \( f(x) \) is said to be an increasing function in \( (a, b) \) if
\( x_1 < x_2 \Rightarrow f(x_1) \le f(x_2) \forall x_1, x_2 \in (a, b) \)
In this way, we can say
\( f(x) \) is increasing in \( (a, b) \) if \( \forall x \in (a, b), f'(x) > 0 \)
Obviously, the angle \( \theta \) made by tangent with +ve direction of x-axis in interval \( (a, b) \) is acute.
\( \Rightarrow \tan \theta \) is +ve \( \Rightarrow \) slope is +ve
\( \Rightarrow \frac{dy}{dx} = f'(x) > 0 \)

Decreasing Function: A function \( f(x) \) is said to be decreasing in the interval \( (a, b) \) if
\( x_1 < x_2 \Rightarrow f(x_1) \ge f(x_2) \forall x_1, x_2 \in (a, b) \)
In other way,
\( f(x) \) is decreasing in interval \( (a, b) \) if \( \forall x \in (a, b), f'(x) < 0 \)
Obviously, the angle \( \theta \) made by tangent with +ve direction of x-axis in interval \( (a, b) \) is obtuse.
\( \Rightarrow \tan \theta \) is –ve \( \Rightarrow \) slope is –ve
\( \Rightarrow \frac{dy}{dx} = f'(x) < 0 \)

[Note: A function \( f(x) \) is said to be:
Strictly increasing if \( x_1 < x_2 \Rightarrow f(x_1) < f(x_2) \forall x_1, x_2 \in (a, b) \)
Strictly decreasing if \( x_1 < x_2 \Rightarrow f(x_1) > f(x_2) \forall x_1, x_2 \in (a, b) \)]

Maximum and Minimum Value of a Function (or Absolute Maximum or Minimum Value)
A function \( f \) is said to attain maximum value at a point \( a \in D_f \), if \( f(a) \ge f(x) \forall x \in D_f \) then \( f(a) \) is called absolute maximum value of \( f \).
A function \( f \) attains minimum value at \( x = b \in D_f \), if \( f(b) \le f(x) \forall x \in D_f \) then \( f(b) \) is called absolute minimum value of \( f \).

Note that a function ‘\( f \)’ may have maximum (or minimum) values in some parts (intervals) of the domain. Such values may occur at more than one point. These are therefore, called local (or relative) maxima (or minima).

Local Maxima and Local Minima (or Relative Extrema)
Local Maxima: A function \( f(x) \) is said to attain a local maxima at \( x = a \), if there exists a neighbourhood \( (a - \delta, a + \delta) \) of ‘\( a \)’ such that \( f(x) < f(a) \forall x \in (a - \delta, a + \delta), x \neq a \), then \( f(a) \) is the local maximum value of \( f(x) \) at \( x = a \).
Local Minima: A function \( f(x) \) is said to attain a local minima at \( x = a \), if there exists a neighbourhood \( (a - \delta, a + \delta) \) of ‘\( a \)’ such that \( f(x) > f(a) \forall x \in (a - \delta, a + \delta), x \neq a \), then \( f(a) \) is called the local minimum value at \( x = a \).

Caution:
(i) A function defined in an interval can reach maximum or minimum values only for those values of \( x \) which lie within the given interval.
(ii) One should not think that the maximum and minimum of a function are its respective largest and smallest values over a given interval.

Test for Identifying Relative (Local) Maxima or Minima
(i) First Derivative Test
Step I: Find \( f'(x) \)
Step II: The equation \( f'(x) = 0 \) is solved to get critical points \( x = c_1, c_2, \dots, c_n \).
Step III: The sign of \( f'(x) \) is studied in the neighbourhood of each critical points. Let one critical point be \( x = c \).
(ii) Second order derivative test
Step I: Find \( f'(x) = 0 \)
Step II: The equation \( f'(x) = 0 \) is solved to get critical points \( x = c_1, c_2, \dots, c_n \).
Step III: \( f''(x) \) is obtained and the sign of \( f''(x) \) is studied for all critical points \( x = c_1, c_2, \dots, c_n \).
Let \( x = c \) be one critical point.

If the sign of \( f'(x) \) changes from +ve to –ve as \( x \) increases through \( c \) (from left to right of \( c \)).
\( x = c \) is relative maxima and \( f(c) \) is relative maximum value.

If the sign of \( f'(x) \) changes from –ve to +ve as \( x \) increases through \( c \) (from left to right of \( c \)).
\( x = c \) is relative minima and \( f(c) \) is relative minimum value.

If \( f''(c) > 0 \)
\( x = c \) is relative minima and \( f(c) \) is relative minimum value.

If \( f''(c) < 0 \)
\( x = c \) is relative maxima and \( f(c) \) is relative maximum value.

If \( f''(c) = 0 \)
Second derivative test fail and first derivative test is used.

Critical point: A point \( x = c \) is called critical point of the function \( f(x) \), if \( f(c) \) exists and either \( f'(c) = 0 \) or \( f'(c) = \infty \) (does not exist).

Point of Inflexion: If \( f(x) \) is a function and \( x = c \) is critical point, then \( x = c \) is called point of inflexion if
(i) \( f'(c) = 0 \) (ii) \( f''(c) = 0 \) (iii) \( f'''(c) \neq 0 \)

SOME USEFUL RESULTS

Area of a square = \( x^2 \) and perimeter = \( 4x \), where \( x \) is side of the square.

Area of a rectangle = \( x \cdot y \), as \( x \) and \( y \) are length and breadth of rectangle and perimeter = \( 2(x + y) \).

Area of a trapezium = \( \frac{1}{2} \) (sum of parallel sides) \( \times \) perpendicular distance between them.

Area of a circle = \( \pi r^2 \), Circumference of a circle = \( 2\pi r \), where \( r \) is the radius.

Volume of sphere = \( \frac{4}{3} \pi r^3 \); Surface area = \( 4\pi r^2 \), where \( r \) is the radius.

Total surface area of a right circular cylinder = \( 2\pi rh + 2\pi r^2 \); Curved surface area = \( 2\pi rh \). Volume = \( \pi r^2 h \), where \( r \) is the radius and \( h \) is the height of the cylinder.

Volume of a right circular cone = \( \frac{1}{3} \pi r^2 h \); Curved surface area = \( \pi rl \), Total surface area = \( \pi r^2 + \pi rl \), where \( r \) is the radius, \( h \) is the height and \( l \) is the slant height of the cone.

Volume of a parallelopiped = \( xyz \) and surface area = \( 2(xy + yz + zx) \), where \( x, y \) and \( z \) are the dimensions of parallelopiped.

Volume of a cube = \( x^3 \) and surface area = \( 6x^2 \), where \( x \) is the side of the cube.

Area of an equilateral triangle = \( \frac{\sqrt{3}}{4} \) (side)\(^2\).

Selected NCERT Questions

Question. A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Answer: Let \( x \) be the radius and \( A \) be the area of the given circle at any time \( t \).
Then \( \frac{dx}{dt} = 5 \) cm/s and \( x = 8 \) cm
Now, \( A = \pi x^2 \Rightarrow \frac{dA}{dt} = 2\pi x \frac{dx}{dt} \Rightarrow \frac{dA}{dt} = 2\pi \times 8 \times 5 = 80\pi \text{ cm}^2/\text{s} \)
Thus, rate of change of area is \( 80\pi \text{ cm}^2/\text{s} \).

Question. Sand is pouring from a pipe at the rate of 12 cm³/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
Answer: Let \( r \) be the radius and \( h \) be the height of the cone so that \( V = \frac{1}{3}\pi r^2 h \)
We have, \( \frac{dV}{dt} = 12\text{ cm}^3\text{/s} \Rightarrow \frac{d}{dt} \left( \frac{1}{3}\pi r^2 h \right) = 12 \) ...(i)
As \( h = \frac{1}{6}r \Rightarrow r = 6h \)
Putting in (i), we get
\( \frac{d}{dt} \left( \frac{1}{3}\pi (6h)^2 \times h \right) = 12 \Rightarrow \frac{d}{dt} (12\pi h^3) = 12 \)
\( \Rightarrow 12\pi \times 3h^2 \frac{dh}{dt} = 12 \Rightarrow \frac{dh}{dt} = \frac{1}{3\pi h^2} \)
when \( h = 4 \) cm, \( \frac{dh}{dt} = \frac{1}{3\pi(4)^2} = \frac{1}{48\pi} \) cm/s

Question. The total cost C(x) in rupees associated with the production of x units of an item is given by \( C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000 \). Find the marginal cost when 17 units are produced.
Answer: Given, \( C(x) = 0.007x^3 - 0.003x^2 + 15x + 4000 \)
\( \therefore \) Marginal cost \( = \frac{d}{dx}C(x) = 0.021x^2 - 0.006x + 15 \)
When \( x = 17 \)
Marginal cost \( = 0.021 \times (17)^2 - 0.006 \times 17 + 15 = Rs 20.967 \)

Question. Find the values of x for which \( y = [x(x - 2)]^2 \) is an increasing function.
Answer: Given, \( y = [x (x - 2)]^2 \)
\( \therefore \frac{dy}{dx} = 2[x(x - 2)] \times (2x - 2) = 4x(x - 1)(x - 2) \)
For increasing function, \( \frac{dy}{dx} > 0 \)
\( 4x(x - 1)(x - 2) > 0 \Rightarrow x(x - 1)(x - 2) > 0 \)
From sign rule,
For \( \frac{dy}{dx} > 0 \) value of \( x = 0 < x < 1 \) and \( x > 2 \)
Therefore, \( y \) is increasing \( \forall x \in (0, 1) \cup (2, \infty) \).

Question. Prove that \( y = \frac{4\sin \theta}{(2 + \cos \theta)} - \theta \) is an increasing function of \( \theta \) in \( [0, \pi/2] \).
Answer: Given, \( y = \frac{4\sin \theta}{2 + \cos \theta} - \theta \)
\( \frac{dy}{d\theta} = \frac{(2 + \cos \theta).4\cos \theta - 4\sin \theta.(0 - \sin \theta)}{(2 + \cos \theta)^2} - 1 \)
\( = \frac{8\cos \theta + 4\cos^2 \theta + 4\sin^2 \theta - (2 + \cos \theta)^2}{(2 + \cos \theta)^2} = \frac{8\cos \theta + 4 - 4 - \cos^2 \theta - 4\cos \theta}{(2 + \cos \theta)^2} \)
\( \Rightarrow \frac{dy}{d\theta} = \frac{4\cos \theta - \cos^2 \theta}{(2 + \cos \theta)^2} = \frac{\cos \theta (4 - \cos \theta)}{(2 + \cos \theta)^2} \)
\( \Rightarrow \frac{dy}{d\theta} = \frac{+ve \times (+ve)}{+ve} \) \( [\because \theta \in [0, \pi/2] \Rightarrow \cos \theta > 0 \text{ and } (4 - \cos \theta) \text{ is +ve as } -1 \le \cos \theta \le 1] \)
\( \Rightarrow \frac{dy}{d\theta} > 0 \)
i.e., \( y = \frac{4\sin \theta}{2 + \cos \theta} - \theta \) is increasing function in \( [0, \pi/2] \).

Question. Prove that the function f given by \( f(x) = \log \sin x \) is strictly increasing on \( (0, \pi/2) \) and strictly decreasing on \( (\pi/2, \pi) \).
Answer: Here, \( f(x) = \log \sin x \Rightarrow f'(x) = \frac{1}{\sin x}(\cos x) = \cot x \)
when \( x \in (0, \pi/2) \) then \( f'(x) > 0 \Rightarrow f(x) \) is strictly increasing on \( (0, \pi/2) \).
when \( x \in (\pi/2, \pi) \) then \( f'(x) < 0 \Rightarrow f(x) \) is strictly decreasing on \( (\pi/2, \pi) \).

Question. Find the equation of tangent to the curve \( y = \frac{x - 7}{(x - 2)(x - 3)} \) at the point, where it cuts the x-axis.
Answer: We have \( y = \frac{x - 7}{(x - 2)(x - 3)} \) ...(i)
Let (i) cuts the x-axis at \( (x, 0) \) i.e., \( y = 0 \)
then \( \frac{x - 7}{(x - 2)(x - 3)} = 0 \Rightarrow x = 7 \)
\( \therefore \) The required point is \( (7, 0) \).
Differentiating equation (i) with respect to \( x \), we get
\( \frac{dy}{dx} = \frac{(x - 2)(x - 3) \cdot 1 - (x - 7) \cdot [(x - 2) + (x - 3)]}{[(x - 2)(x - 3)]^2} \)
\( = \frac{x^2 - 5x + 6 - (x - 7)(2x - 5)}{(x^2 - 5x + 6)^2} = \frac{x^2 - 5x + 6 - 2x^2 + 19x - 35}{(x^2 - 5x + 6)^2} = \frac{-x^2 + 14x - 29}{(x^2 + 6 - 5x)^2} \)
\( \frac{dy}{dx} \bigg|_{(7, 0)} = \frac{-49 + 98 - 29}{(49 - 35 + 6)^2} = \frac{20}{400} = \frac{1}{20} \)
\( \therefore \) Equation of tangent is \( y - y_1 = m(x - x_1) \)
\( \Rightarrow y - 0 = \frac{1}{20}(x - 7) \) or \( x - 20y - 7 = 0 \)

Question. Find the equations of the tangent and normal to the curve \( x = a \sin^3 \theta \) and \( y = a \cos^3 \theta \) at \( \theta = \frac{\pi}{4} \).
Answer: Given, \( x = a \sin^3 \theta \) and \( y = a \cos^3 \theta \)
\( \Rightarrow \frac{dx}{d\theta} = 3a \sin^2 \theta \cos \theta \) and \( \frac{dy}{d\theta} = -3a \cos^2 \theta \sin \theta \)
\( \Rightarrow \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-3a \cos^2 \theta \sin \theta}{3a \sin^2 \theta \cos \theta} = -\cot \theta \)
\( \Rightarrow \) Slope of tangent to the given curve at \( \theta = \frac{\pi}{4} \) is \( \frac{dy}{dx} \bigg|_{\theta = \frac{\pi}{4}} = -\cot \frac{\pi}{4} = -1 \).
Since for \( \theta = \frac{\pi}{4}, x = a \sin^3 \frac{\pi}{4} \) and \( y = a \cos^3 \frac{\pi}{4} \)
\( \Rightarrow x = a\left(\frac{1}{\sqrt{2}}\right)^3 \) and \( y = a\left(\frac{1}{\sqrt{2}}\right)^3 \Rightarrow x = \frac{a}{2\sqrt{2}} \) and \( y = \frac{a}{2\sqrt{2}} \)
i.e., co-ordinates of the point of contact \( = \left(\frac{a}{2\sqrt{2}}, \frac{a}{2\sqrt{2}}\right) \)
\( \therefore \) Equation of tangent is \( \left(y - \frac{a}{2\sqrt{2}}\right) = (-1) \cdot \left(x - \frac{a}{2\sqrt{2}}\right) \Rightarrow y - \frac{a}{2\sqrt{2}} = -x + \frac{a}{2\sqrt{2}} \Rightarrow x + y = \frac{a}{\sqrt{2}} \)
Also slope of normal (at \( \theta = \frac{\pi}{4} \)) \( = -\frac{1}{\text{slope of tangent}} = -\frac{1}{-1} = 1 \)
\( \therefore \) Equation of normal is \( \left(y - \frac{a}{2\sqrt{2}}\right) = (1) \cdot \left(x - \frac{a}{2\sqrt{2}}\right) \Rightarrow y - \frac{a}{2\sqrt{2}} = x - \frac{a}{2\sqrt{2}} \Rightarrow y - x = 0 \)

Question. Find the slope of the normal to the curve \( x = 1 - a \sin \theta, y = b \cos^2 \theta \) at \( \theta = \frac{\pi}{2} \).
Answer: Here \( x = 1 - a \sin \theta \) and \( y = b \cos^2 \theta \)
Differentiating both sides w.r.t. \( \theta \), we have
\( \frac{dx}{d\theta} = -a \cos \theta, \frac{dy}{d\theta} = -2b \sin \theta \cos \theta \)
Now, \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-2b \sin \theta \cos \theta}{-a \cos \theta} = \frac{2b \sin \theta}{a} \)
Slope of normal at \( \theta = \frac{\pi}{2} \) is \( \frac{-1}{\left[ \frac{dy}{dx} \right]_{\theta = \frac{\pi}{2}}} = \frac{-1}{\frac{2b}{a} \sin \frac{\pi}{2}} = \frac{-a}{2b \sin \frac{\pi}{2}} = -\frac{a}{2b} \)
Thus, slope of normal at \( \theta = \frac{\pi}{2} \) is \( -\frac{a}{2b} \).

Question. Find the point on the curve \( y = x^3 - 11x + 5 \) at which the equation of tangent is \( y = x - 11 \).
Answer: Let the required point of contact be \( (x_1, y_1) \).
Given curve is \( y = x^3 - 11x + 5 \) ...(i)
\( \therefore \frac{dy}{dx} = 3x^2 - 11 \Rightarrow \left[ \frac{dy}{dx} \right]_{(x_1, y_1)} = 3x_1^2 - 11 \)
i.e., slope of tangent at \( (x_1, y_1) \) to give curve (i) \( = 3x_1^2 - 11 \)
From question \( 3x_1^2 - 11 = \) slope of line \( y = x - 11 \), which is also tangent.
i.e., \( 3x_1^2 - 11 = 1 \Rightarrow 3x_1^2 = 12 \Rightarrow x_1^2 = 4 \Rightarrow x_1 = \pm 2 \)
Since \( (x_1, y_1) \) lie on curve (i)
\( \therefore y_1 = x_1^3 - 11x_1 + 5 \)
When \( x_1 = 2, y_1 = 2^3 - 11 \times 2 + 5 = 8 - 22 + 5 = -9 \)
When \( x_1 = -2, y_1 = (-2)^3 - 11 \times (-2) + 5 = -8 + 22 + 5 = 19 \)
But \( (-2, 19) \) does not satisfy the line \( y = x - 11 \).
Therefore \( (2, -9) \) is the required point of curve at which tangent is \( y = x - 11 \).

Question. Find the equation of the tangent line of the curve \( y = x^2 - 2x + 7 \) which is (a) parallel to the line \( 2x - y + 9 = 0 \), (b) perpendicular to the line \( 5y - 15x = 13 \).
Answer: Here, \( y = x^2 - 2x + 7 \)
\( \frac{dy}{dx} = 2x - 2 = 2(x - 1) \)
(a) Slope of the line \( 2x - y + 9 = 0 \) is \( -\frac{\text{Coefficient of } x}{\text{Coefficient of } y} = -\frac{2}{-1} = 2 \)
It is given that tangent is parallel to the line.
\( \therefore 2(x - 1) = 2 \Rightarrow x - 1 = 1 \Rightarrow x = 2 \)
when \( x = 2 \) then \( y = (2)^2 - 2 \times 2 + 7 = 4 - 4 + 7 = 7 \)
\( \therefore \) Equation of tangent at \( (2, 7) \) is \( y - 7 = 2(x - 2) \Rightarrow y - 7 = 2x - 4 \Rightarrow 2x - y + 3 = 0 \)
(b) Slope of line \( 5y - 15x = 13 \) is \( -\frac{\text{Coefficient of } x}{\text{Coefficient of } y} = -\frac{(-15)}{5} = 3 \)
It is given that tangent is perpendicular to the line \( 5y - 15x = 13 \)
\( \therefore 2(x - 1) \times 3 = -1 \Rightarrow 6x - 6 = -1 \Rightarrow 6x = 5 \Rightarrow x = 5/6 \)
when \( x = 5/6 \) then \( y = \left(\frac{5}{6}\right)^2 - 2 \times \frac{5}{6} + 7 = \frac{25}{36} - \frac{5}{3} + 7 = \frac{25 - 60 + 252}{36} = \frac{217}{36} \)
\( \therefore \) Equation of tangent at \( \left(\frac{5}{6}, \frac{217}{36}\right) \) is \( y - \frac{217}{36} = -\frac{1}{3}\left(x - \frac{5}{6}\right) \Rightarrow \frac{36y - 217}{36} = \frac{-6x + 5}{18} \)
\( \Rightarrow 36y - 217 = -12x + 10 \Rightarrow 12x + 36y - 227 = 0 \)

Question. Prove that the curves \( x = y^2 \) and \( xy = k \) cut at right angles if \( 8k^2 = 1 \).
Answer: As we know, two curves intersect at right angles if the tangents to the curves at the point of intersection are perpendicular to each other, i.e., product of slope of these two curves is -1.
We have \( x = y^2 \)
Differentiating with respect to \( x \), we have \( 1 = 2y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{2y} = m_1 \), (let) ...(i)
\( xy = k \)
Also, differentiating with respect to \( x \), we have \( x \cdot \frac{dy}{dx} + y \cdot 1 = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x} = m_2 \), (let) ...(ii)
On solving the equations of the two curves
\( xy = k \) and \( x = y^2 \)
We get \( x = k^{2/3} \) and \( y = k^{1/3} \)
Putting these values in (i) and (ii), we have \( m_1 = \frac{1}{2k^{1/3}} \) and \( m_2 = \frac{-k^{1/3}}{k^{2/3}} = -k^{-1/3} \)
For the curves to intersect at right angles \( m_1 \times m_2 = -1 \Rightarrow \frac{1}{2k^{1/3}} \times (-k^{-1/3}) = -1 \Rightarrow \frac{1}{2k^{2/3}} = 1 \)
\( \left(\frac{1}{2}\right)^3 = (k^{2/3})^3 \Rightarrow \frac{1}{8} = k^2 \Rightarrow 8k^2 = 1 \)
Hence, the result is proved.

Question. Find the equation of the tangent and normal to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) at the point \( (x_0, y_0) \).
Answer: Here, equation of the given curve is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
Differentiating both sides w.r.t. \( x \), we have
\( \frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{2x}{a^2} \times \frac{b^2}{2y} = \frac{b^2 x}{a^2 y} \)
\( \Rightarrow \left(\frac{dy}{dx}\right) \) at \( (x_0, y_0) = \frac{b^2 x_0}{a^2 y_0} \)
\( \therefore \) Equation of tangent at \( (x_0, y_0) \) is \( y - y_0 = \frac{b^2 x_0}{a^2 y_0}(x - x_0) \Rightarrow a^2 y y_0 - a^2 y_0^2 = b^2 x x_0 - b^2 x_0^2 \)
\( \Rightarrow b^2 x x_0 - a^2 y y_0 = b^2 x_0^2 - a^2 y_0^2 \Rightarrow \frac{xx_0}{a^2} - \frac{yy_0}{b^2} = \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} \)
\( \Rightarrow \frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1 \) \( \left[ \because \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} = 1 \right] \)
Equation of normal at \( (x_0, y_0) \) is \( y - y_0 = -\frac{1}{b^2 x_0/a^2 y_0}(x - x_0) \Rightarrow y - y_0 = -\frac{a^2 y_0}{b^2 x_0}(x - x_0) \)
\( \Rightarrow \frac{y - y_0}{a^2 y_0} = -\frac{x - x_0}{b^2 x_0} \Rightarrow \frac{a^2 y_0}{y - y_0} + \frac{b^2 x_0}{x - x_0} = 0 \)

Question. Find the local maximum or minimum if any of the function \( f(x) = \sin x - \cos x, 0 < x < 2\pi \). Also, find the local extrema values.
Answer: \( f'(x) = \cos x + \sin x = 0 \Rightarrow \tan x = -1 \Rightarrow x = \frac{3\pi}{4}, \frac{7\pi}{4} \)
At \( x = \frac{3\pi}{4}, f'(x) = \cos x + \sin x \),
when \( x < \frac{3\pi}{4}, f'(x) = +ve \)
when \( x > \frac{3\pi}{4}, f'(x) = -ve \Rightarrow x = \frac{3\pi}{4} \) is a point of local maxima.
At \( x = \frac{7\pi}{4}, f'(x) = \cos x + \sin x \)
when \( x < \frac{7\pi}{4}, f'(x) = -ve \)
when \( x > \frac{7\pi}{4}, f'(x) = +ve \Rightarrow x = \frac{7\pi}{4} \) is a point of local minima.
\( f\left(\frac{3\pi}{4}\right) = \sin \frac{3\pi}{4} - \cos \frac{3\pi}{4} = \frac{1}{\sqrt{2}} - \left(-\frac{1}{\sqrt{2}}\right) = \sqrt{2} \) is the local maximum value.
\( f\left(\frac{7\pi}{4}\right) = \sin \frac{7\pi}{4} - \cos \frac{7\pi}{4} = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} \) is the local minimum value.

Question. Find the absolute maximum value and the absolute minimum value of the following function in the given interval. \( f(x) = 4x - \frac{1}{2}x^2, x \in \left[-2, \frac{9}{2}\right] \)
Answer: Here, \( f(x) = 4x - \frac{1}{2}x^2, x \in \left[-2, \frac{9}{2}\right] \Rightarrow f'(x) = 4 - x \)
For maximum or minimum \( f'(x) = 0 \Rightarrow 4 - x = 0 \Rightarrow x = 4 \)
\( f(-2) = 4(-2) - \frac{1}{2}(-2)^2 = -8 - 2 = -10 \)
\( f(4) = 4(4) - \frac{1}{2}(4)^2 = 16 - 8 = 8 \)
\( f\left(\frac{9}{2}\right) = 4\left(\frac{9}{2}\right) - \frac{1}{2}\left(\frac{9}{2}\right)^2 = 18 - \frac{81}{8} = \frac{63}{8} \)
Thus, absolute maximum value is 8 at \( x = 4 \) and absolute minimum value is -10 at \( x = -2 \).

Question. From first equation of the curve, we have \( 3x^2 - 3y^2 - 6xy \frac{dy}{dx} = 0 \)
\( \Rightarrow \frac{dy}{dx} = \frac{x^2 - y^2}{2xy} = (m_1) \) say and second equation of the curve gives \( 6xy + 3x^2 \frac{dy}{dx} - 3y^2 \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{-2xy}{x^2 - y^2} = (m_2) \) say
Since \( m_1 \cdot m_2 = -1 \). Therefore, correct answer is (b).
Answer: (b)

Question. \( \frac{dy}{dx} = \cos x \). Therefore, slope of normal \( = \left( \frac{-1}{\cos x} \right)_{x=0} = -1 \).
Hence, the equation of normal is \( y - 0 = -1 (x - 0) \) or \( x + y = 0 \). Therefore, correct answer is (c).
Answer: (c)

Fill in the Blanks

Question. If the radius of the circle is increasing at the rate of 0.5 cm/s, then the rate of increase of its circumference is _____________ .
Answer: Given, \( \frac{dr}{dt} = 0.5 \text{ cm/s} \)
\( \therefore \frac{dC}{dt} = \frac{d(2\pi r)}{dt} = 2\pi \frac{dr}{dt} = 2\pi \times 0.5 \text{ cm/s} = \pi \text{ cm/sec.} \)

Question. The values of a for which the function \( f(x) = \sin x - ax + b \) increases on R are _____________ .
Answer: \( (-\infty, -1) \)

Question. The rate of change of volume of a sphere with respect to its surface area, when radius is 2 cm, is _____________ .
Answer: \( 1 \text{ cm}^3/\text{cm}^2 \)

Question. The equation of the tangent to the curve \( y = \sec x \) at the point (0, 1) is _____________ .
Answer: We have, slope of the tangent to the curve \( y = \sec x \) is given by \( \frac{dy}{dx} = \sec x \cdot \tan x \)
\( \therefore \) Slope of tangent at \( (0, 1) = \sec 0 \times \tan 0 = 0 \)
\( \therefore \) Equation of tangent at the point (0, 1) be \( \frac{y - 1}{x - 0} = 0 \Rightarrow y - 1 = 0 \Rightarrow y = 1 \)

Question. The least value of the function \( f(x) = ax + \frac{b}{x} \), \( (a > 0, b > 0, x > 0) \) is _____________ .
Answer: \( 2\sqrt{ab} \)

Very Short Answer Questions

Question. Find the slope of the tangent to the curve \( y = 2 \sin^2 (3x) \) at \( x = \frac{\pi}{6} \).
Answer: We have, \( y = 2 \sin^2 (3x) \)
\( \Rightarrow \frac{dy}{dx} = 2 \times 2 \sin (3x) \times \cos (3x) \times 3 \)
\( \Rightarrow \frac{dy}{dx} = 6 \sin (6x) \)
\( \therefore \) Slope of tangent at \( x = \frac{\pi}{6} \) is \( 6 \sin \left( 6 \times \frac{\pi}{6} \right) = 6 \sin \pi = 6 \times 0 = 0 \)

Question. Find the rate of change of the area of a circle with respect to its radius ‘r’ when r = 4 cm.
Answer: If A is area and r is the radius of a circle, then \( A = \pi r^2 \Rightarrow \frac{dA}{dr} = 2\pi r \)
\( \therefore \left[ \frac{dA}{dr} \right]_{r=4} = 8\pi \text{ cm}^2/\text{cm} \)

Question. An edge of a variable cube is increasing at the rate of 5 cm per second. How fast is the volume increasing when the side is 15 cm?
Answer: Let x be the edge of the cube and V be the volume of the cube at any time t.
Given, \( \frac{dx}{dt} = 5 \text{ cm/s} \), \( x = 15 \text{ cm} \)
Since we know the volume of cube \( = (\text{side})^3 \) i.e., \( V = x^3 \).
\( \Rightarrow \frac{dV}{dt} = 3x^2 \cdot \frac{dx}{dt} \Rightarrow \frac{dV}{dt} = 3 \cdot (15)^2 \times 5 = 3375 \text{ cm}^3 / \text{sec} \)

Question. Find the slope of the tangent to the curve \( x = t^2 + 3t - 8, y = 2t^2 - 2t - 5 \) at t = 2 .
Answer: Slope of the tangent \( = \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t - 2}{2t + 3} \)
\( \Rightarrow \frac{dy}{dx} \bigg|_{t=2} = \frac{4(2) - 2}{2(2) + 3} = \frac{6}{7} \)

Question. If \( y = \log_e x \), then find \( \Delta y \) when x = 3 and \( \Delta x = 0.03 \).
Answer: We have, \( y = \log_e x \)
\( \therefore \Delta y = \frac{dy}{dx} \cdot \Delta x = \frac{1}{x} \cdot \Delta x = \frac{1}{3} \cdot 0.03 = 0.01 \)